Chapter 4 of 7

All 7 chapters
  1. 1Exponential functions, readunderstood
  2. 2Logarithms, readunderstood
  3. 3Laws of logarithms, readunderstood
  4. 4Equations with powers, readunderstood
  5. 5The number e and natural logs, readunderstood
  6. 6Growth and decay models, readunderstood
  7. 7Straight lines from logs, readunderstood

M5.2.4 Chapter 4

Equations with powers

How do we solve an equation where the unknown is in a power, such as 3ˣ = 20, when trying numbers is slow?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Solve 3ˣ = 20 by taking logs
  • Solve 4ˣ − 5(2ˣ) + 4 = 0 as a quadratic in 2ˣ

In chapter 2 you solved 2ˣ = 64 by counting powers. For 3ˣ = 20 that fails, because 20 is not a power of 3. We need a tool that brings the power x down to the ground. The power law of logarithms does exactly this: log (aˣ) = x log a.

The method for aˣ = b has three moves. First take logs of both sides. This is allowed because both sides are above 0, and two equal numbers have equal logs. Second use the power law to bring x down. Third divide by log a. The answer is x = log b ÷ log a, with the logs to any one base. A calculator log key uses base 10.

Worked example

Solve 3ˣ = 20. Give x to 3 significant figures.

  1. Estimate first. 3² = 9 and 3³ = 27, and 20 is between them. So x is between 2 and 3.
  2. Take logs of both sides: log (3ˣ) = log 20.
  3. The power law brings x down: x log 3 = log 20.
  4. Divide both sides by log 3: x = log 20 ÷ log 3.
  5. Use log 20 = 1.30103 and log 3 = 0.47712. Then x = 1.30103 ÷ 0.47712 = 2.7268.
  6. Round once, at the end. To 3 significant figures, x = 2.73. This agrees with the estimate.

Answer: x = 2.73

A graph with the x axis x and the y axis y. A curve for y = 3ˣ rises from the point (0, 1). A dashed horizontal line named y = 20 crosses it at the point (2.73, 20). Dashed guide lines from that point meet the axes. The point is named (2.73, 20).(2.73, 20)y = 2012341020300xy
The solution of 3ˣ = 20 is the x value where the curve y = 3ˣ meets the line y = 20.

Often the unknown is in a longer power, or in a power on both sides. The moves are the same. Take logs, bring the powers down, and collect every term in x on one side. Keep the brackets: a power such as 2x − 1 must be multiplied by log 3 as a whole.

Worked example

Solve 3²ˣ⁻¹ = 7ˣ. Give x to 3 significant figures.

  1. Take logs of both sides: log (3²ˣ⁻¹) = log (7ˣ).
  2. Bring the powers down. The whole power 2x − 1 multiplies log 3: (2x − 1) log 3 = x log 7.
  3. Expand the bracket: 2x log 3 − log 3 = x log 7.
  4. Collect the x terms on one side: 2x log 3 − x log 7 = log 3.
  5. Factorise: x (2 log 3 − log 7) = log 3. So x = log 3 ÷ (2 log 3 − log 7).
  6. Use log 3 = 0.47712 and log 7 = 0.84510. Then 2 log 3 − log 7 = 0.95424 − 0.84510 = 0.10914.
  7. Divide: x = 0.47712 ÷ 0.10914 = 4.3716. To 3 significant figures, x = 4.37.

Answer: x = 4.37

Be careful with an inequality. A log to a base above 1 keeps the order, so if one number is bigger, its log is bigger. But the log of a number between 0 and 1 is negative. When you divide both sides by a negative number, the inequality sign flips.

Worked example

A sample has a mass of 1 kg and halves every hour. After x hours its mass is 0.5ˣ kg. Find when the mass is below 0.1 kg. Give the smallest whole number of hours.

  1. We need 0.5ˣ < 0.1.
  2. Take logs of both sides: x log 0.5 < log 0.1.
  3. Now log 0.5 = −0.30103 and log 0.1 = −1. Both are negative.
  4. Divide by log 0.5, which is negative. The sign flips: x > log 0.1 ÷ log 0.5.
  5. To 3 significant figures, x > −1 ÷ −0.30103 = 3.32.
  6. The smallest whole number above 3.32 is 4. Check: 0.5⁴ = 0.0625, which is below 0.1, and 0.5³ = 0.125, which is not.

Answer: x > 3.32, so the smallest whole number of hours is 4.

A graph with the x axis hours and the y axis mass (kg). A curve for the mass falls from the point (0, 1) towards the x axis. A dashed horizontal line named y = 0.1 meets it at the point A, (3.32, 0.1). Dashed guide lines from that point meet the axes. To the right of that point the curve is below the line.Ay = 0.11234560.510hoursmass (kg)
The mass is below 0.1 kg to the right of the point A, (3.32, 0.1), where the curve is under the line.

Some equations look new but are a quadratic in disguise. In 4ˣ − 5(2ˣ) + 4 = 0 the base 4 is 2². So 4ˣ = (2²)ˣ = (2ˣ)². Both powers are now powers of 2ˣ. Write u for 2ˣ and the equation becomes a quadratic in u. Solve for u, then go back to x. One fact decides which roots you keep: 2ˣ is always above 0, so u must be above 0.

Worked example

Solve 4ˣ − 5(2ˣ) + 4 = 0.

  1. Write 4ˣ as (2ˣ)². Let u = 2ˣ. The equation becomes u² − 5u + 4 = 0.
  2. Factorise: (u − 1)(u − 4) = 0. So u = 1 or u = 4.
  3. Both are above 0, so both can be 2ˣ. Go back to x.
  4. Because 2⁰ = 1, the equation 2ˣ = 1 gives x = 0.
  5. Because 2² = 4, the equation 2ˣ = 4 gives x = 2.
  6. Check x = 0: 4⁰ − 5(2⁰) + 4 = 1 − 5 + 4 = 0. Check x = 2: 16 − 20 + 4 = 0. Both work.

Answer: x = 0 or x = 2.

A graph with the x axis x and the y axis y. A curve for y = 4ˣ − 5(2ˣ) + 4 comes down from the upper left, crosses the x axis at the point (0, 0), dips below the axis, then crosses it again at the point (2, 0) and climbs steeply. The two crossing points are named A, at (0, 0), and B, at (2, 0).AB−1123−4480xy
The curve y = 4ˣ − 5(2ˣ) + 4 meets the x axis at A, (0, 0), and B, (2, 0). These are the two solutions, x = 0 and x = 2.

The power may be shifted, as in 2²ˣ⁺¹. Split it with the law of indices: 2²ˣ⁺¹ = 2²ˣ × 2¹ = 2 × (2ˣ)². The 2 in front changes the quadratic, so it may not factorise as simply. The same rules apply: u = 2ˣ must be above 0.

Worked example

Solve 2²ˣ⁺¹ − 5(2ˣ) + 2 = 0.

  1. Split the power: 2²ˣ⁺¹ = 2²ˣ × 2¹ = 2 × (2ˣ)².
  2. Let u = 2ˣ. The equation becomes 2u² − 5u + 2 = 0.
  3. Factorise: (2u − 1)(u − 2) = 0. So u = 1/2 or u = 2.
  4. Both are above 0, so both can be 2ˣ.
  5. Because 2⁻¹ = 1/2, the equation 2ˣ = 1/2 gives x = −1.
  6. Because 2¹ = 2, the equation 2ˣ = 2 gives x = 1.
  7. Check x = 1: 2³ − 5(2) + 2 = 8 − 10 + 2 = 0. Check x = −1: 2⁻¹ − 5(2⁻¹) + 2 = 0.5 − 2.5 + 2 = 0.

Answer: x = −1 or x = 1.

Worked example

Solve 9ˣ − 4(3ˣ) − 5 = 0. Give any answer that is not exact to 3 significant figures.

  1. Write 9ˣ as (3ˣ)². Let u = 3ˣ. Then u² − 4u − 5 = 0.
  2. Factorise: (u − 5)(u + 1) = 0. So u = 5 or u = −1.
  3. Now u = 3ˣ, and 3ˣ is always above 0. So u = −1 has no solution for x.
  4. Only u = 5 is left: 3ˣ = 5.
  5. Take logs: x log 3 = log 5, so x = log 5 ÷ log 3 = 0.69897 ÷ 0.47712 = 1.46498.
  6. To 3 significant figures, x = 1.46, rounded once from 1.46498. There is only one solution.

Answer: x = 1.46

Equations with logs in them need a different set of checks. Use the laws to write one log, change to index form, and solve. Then test every answer: each number inside an original log must be above 0. An answer that makes a number inside a log zero or negative is an extraneous root, and it must be thrown away.

Worked example

Solve log₂ x + log₂ (x − 2) = 3.

  1. Both logs are to the base 2, so use the product law: log₂ [x(x − 2)] = 3.
  2. Change to index form: x(x − 2) = 2³ = 8.
  3. Expand and collect: x² − 2x − 8 = 0.
  4. Factorise: (x − 4)(x + 2) = 0. So x = 4 or x = −2.
  5. Test the answers in the original equation. We need x above 0 and x − 2 above 0, so x must be above 2.
  6. The value x = −2 breaks both rules, so reject it. For x = 4: log₂ 4 + log₂ 2 = 2 + 1 = 3. So x = 4.

Answer: x = 4

Three more kinds of equation with logs come up in exams. When the bases differ, change them to one base. When a log appears as a square and on its own, call it y. When two logs are tied in a pair of equations, use the product and quotient laws to turn the pair into two ordinary equations. In every case, test each answer against the rule that numbers inside a log must be above 0.

Worked example

Solve log₂ x = log₄ (x + 6).

  1. The bases differ, so change log₄ to base 2: log₄ (x + 6) = log₂ (x + 6) ÷ log₂ 4.
  2. Since 2² = 4, log₂ 4 = 2. So log₄ (x + 6) = 1/2 log₂ (x + 6).
  3. The equation becomes log₂ x = 1/2 log₂ (x + 6). Multiply both sides by 2: 2 log₂ x = log₂ (x + 6).
  4. The power law makes the left side log₂ (x²). So log₂ (x²) = log₂ (x + 6).
  5. Equal logs to the same base mean equal numbers inside: x² = x + 6, so x² − x − 6 = 0.
  6. Factorise: (x − 3)(x + 2) = 0. So x = 3 or x = −2.
  7. We need x above 0 for log₂ x. So reject x = −2. Check x = 3: log₂ 3 = 1.585 and log₄ 9 = 1.585. So x = 3.

Answer: x = 3

Worked example

Solve (log₂ x)² − 3 log₂ x + 2 = 0.

  1. The log appears as a square and on its own. Let y = log₂ x. The equation becomes y² − 3y + 2 = 0.
  2. Factorise: (y − 1)(y − 2) = 0. So y = 1 or y = 2.
  3. Go back to x. If log₂ x = 1, then x = 2¹ = 2.
  4. If log₂ x = 2, then x = 2² = 4.
  5. Both are above 0, so both are allowed. Check x = 4: (log₂ 4)² − 3 log₂ 4 + 2 = 4 − 6 + 2 = 0.

Answer: x = 2 or x = 4.

Worked example

Solve log₃ x + log₃ y = 4 and log₃ x − log₃ y = 2 together.

  1. Use the product law on the first equation: log₃ (xy) = 4. So xy = 3⁴ = 81.
  2. Use the quotient law on the second: log₃ (x ÷ y) = 2. So x ÷ y = 3² = 9, which gives x = 9y.
  3. Put x = 9y into xy = 81: 9y² = 81. So y² = 9.
  4. Then y = 3 or y = −3. A log needs y above 0, so y = 3.
  5. Then x = 9 × 3 = 27. Check: log₃ 27 + log₃ 3 = 3 + 1 = 4, and log₃ 27 − log₃ 3 = 3 − 1 = 2.

Answer: x = 27 and y = 3.

See it three ways

With things

A pile of 1 seed doubles every day: 1, 2, 4, 8, 16. The pile passes 12 seeds between day 3 (8 seeds) and day 4 (16 seeds).

As a picture

A graph with the x axis days and the y axis seeds. A curve for the number of seeds rises from the point (0, 1). A dashed horizontal line named y = 12 meets it at the point (3.58, 12), named (3.58, 12). Dashed guide lines from that point meet the axes.(3.58, 12)y = 1212345481216200daysseeds

In words

The pile reaches 12 seeds after a bit more than 3.5 days. The graph shows where the curve meets the line at 12, and logs give that point exactly.

In symbols

2ˣ = 12, so x = log 12 ÷ log 2 = 1.07918 ÷ 0.30103 = 3.58496, which is 3.58 to 2 decimal places.

A table with the heads equation and first move. Rows: aˣ = b and take logs; quadratic in aˣ and let u = aˣ; log x + log y and one log.equationfirst moveaˣ = btake logsquadratic in aˣlet u = aˣlog x + log yone log
Three kinds of equation in this chapter and the first move for each.

Try it

Check a solution on a calculator

You need: a scientific calculator, paper and a pencil.

  1. Work out 3 to the power 2.73 with the power key.
  2. Work out 3 to the power 2.727 and 3 to the power 2.7268.
  3. Write the three results to 2 decimal places.

What you will see: The results get closer to 20 as the power gets closer to the exact value of x. A solution found by logs can always be checked this way. Put it back in the equation and see if you get 20.

If you cannot try it, this is what happens.
A table with the heads power and value. Rows: 2.73 and 20.07; 2.727 and 20.00; 2.7268 and 20.00.powervalue2.7320.072.72720.002.726820.00

A wrong idea many people have

To solve 3ˣ = 20, divide: x = 20 ÷ 3. Or write x = log (20 ÷ 3).

Why it feels right: In 3x = 20 you divide by 3. And the log of a quotient is a familiar law, so a log with 20 ÷ 3 inside looks tidy.

The test that breaks it: If x = 20 ÷ 3 = 6.67, then 3ˣ is about 1 500, not 20. If x = log (20 ÷ 3) = 0.824, then 3ˣ is about 2.5, not 20. Neither works. The correct x = 2.73 gives 20.

What is true: Here x is a power, so division by 3 does not undo it. The log undoes it: x = log 20 ÷ log 3, a quotient of two logs. The quotient law of logs is about a difference, not this. Catch the slip by putting your answer back into 3ˣ. Also keep two more figures than you need until the last line, and round once.

A wrong idea many people have

Both roots of the quadratic in u always give a solution for x.

Why it feels right: Both roots solve the quadratic. Handing back only one feels like losing half the answer.

The test that breaks it: In 9ˣ − 4(3ˣ) − 5 = 0 the roots are u = 5 and u = −1. Try 3ˣ = −1. No power of 3 is negative, so there is no real x, and a log of −1 does not exist.

What is true: A root for u must satisfy u = aˣ, which is above 0. Throw away any root with u ≤ 0. The same rule applies to a log equation: reject any answer that puts zero or a negative number inside a log.

A wrong idea many people have

Solving 4ˣ − 5(2ˣ) + 4 = 0 gives only x = 2, because x = 0 is too plain to count.

Why it feels right: The root u = 4 gives the clear answer 2ˣ = 4, x = 2. The root u = 1 gives x = 0, which does not look like a power.

The test that breaks it: Put x = 0 into the equation: 4⁰ − 5(2⁰) + 4 = 1 − 5 + 4 = 0. It works. So x = 0 is a solution, and giving only x = 2 loses it.

What is true: A quadratic in u has two roots, and each positive root gives an x. The value 2⁰ = 1 means x = 0 is a normal answer. Catch the slip by solving for every root of u before going back to x, then checking each in the original equation.

A wrong idea many people have

Dividing by a negative log does not change the inequality sign.

Why it feels right: For positive numbers the sign never changes, and the log sign is easy to forget.

The test that breaks it: For 0.5ˣ < 0.1, take logs: x log 0.5 < log 0.1, where log 0.5 is negative. Without a flip, x < 3.32, and x = 1 would pass. But 0.5¹ = 0.5, which is not below 0.1. With the flip, x > 3.32, and x = 4 gives 0.0625.

What is true: Dividing both sides of an inequality by a negative number reverses it. A base between 0 and 1 gives a negative log. Catch the slip by testing a value of x on each side of your boundary.

Key words

take logs
apply a log to both sides of an equation, to bring a power down with the power law.
quadratic in disguise
an equation with two powers of the same kind, such as 4ˣ and 2ˣ, that becomes a quadratic after a substitution.
extraneous root
an answer found by correct algebra that fails the original equation, such as a value that puts a negative number inside a log.

Back to your guess

The answer is B: x is between 2 and 3.

Since 3² = 9 and 3³ = 27, and 20 lies between 9 and 27, the power x lies between 2 and 3. The value 20 ÷ 3 is wrong because x is a power, not a number to divide by. To find x exactly we take logarithms, which this chapter shows step by step.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which of these is the solution of 5ˣ = 12?

    Tap your answer.

    1. Not this one. x is a power, and dividing by 5 does not undo a power. 5 to the power 2.4 is about 48, not 12. Go to M5.2.4. Try another.

    2. Not this one. The log of a quotient is a difference of logs, log 12 − log 5. It is not the quotient of two logs. Check: 5 to the power 0.38 is about 1.8, not 12. Go to M5.2.4. Try another.

    3. Yes, that is right. Taking logs gives x log 5 = log 12. Divide by log 5 to get x = log 12 ÷ log 5, about 1.54.

    4. Not this one. You turned the fraction upside down. Since 5¹ = 5 is below 12, x must be above 1. This answer is below 1. Go to M5.2.4. Try another.

    See the answer

    x = log(12) ÷ log(5)

    Taking logs gives x log 5 = log 12. Divide by log 5 to get x = log 12 ÷ log 5, about 1.54.

    • If you chose A: x is a power, and dividing by 5 does not undo a power. 5 to the power 2.4 is about 48, not 12. Go to M5.2.4.
    • If you chose B: The log of a quotient is a difference of logs, log 12 − log 5. It is not the quotient of two logs. Check: 5 to the power 0.38 is about 1.8, not 12. Go to M5.2.4.
    • If you chose D: You turned the fraction upside down. Since 5¹ = 5 is below 12, x must be above 1. This answer is below 1. Go to M5.2.4.

  2. Solve 16ˣ − 20(4ˣ) + 64 = 0.

    Tap your answer.

    1. Yes, that is right. Let u = 4ˣ. Then 16ˣ = u², and u² − 20u + 64 = 0, which is (u − 4)(u − 16) = 0. So 4ˣ = 4 gives x = 1, and 4ˣ = 16 gives x = 2.

    2. Not this one. The root u = 4 gives x = 1. You have lost a solution. Each positive root of u gives a value of x. Go to M5.2.4. Try another.

    3. Not this one. Those come from factorising with the wrong signs, (u + 4)(u + 16). The quadratic is u² − 20u + 64, so the roots are positive. Go to M5.2.4. Try another.

    4. Not this one. These are the values of u = 4ˣ. You must go back to x: 4ˣ = 4 gives x = 1 and 4ˣ = 16 gives x = 2. Go to M5.2.4. Try another.

    See the answer

    x = 1 or x = 2

    Let u = 4ˣ. Then 16ˣ = u², and u² − 20u + 64 = 0, which is (u − 4)(u − 16) = 0. So 4ˣ = 4 gives x = 1, and 4ˣ = 16 gives x = 2.

    • If you chose B: The root u = 4 gives x = 1. You have lost a solution. Each positive root of u gives a value of x. Go to M5.2.4.
    • If you chose C: Those come from factorising with the wrong signs, (u + 4)(u + 16). The quadratic is u² − 20u + 64, so the roots are positive. Go to M5.2.4.
    • If you chose D: These are the values of u = 4ˣ. You must go back to x: 4ˣ = 4 gives x = 1 and 4ˣ = 16 gives x = 2. Go to M5.2.4.

  3. Solve log₃ x + log₃ (x − 8) = 2.

    Tap your answer.

    1. Not this one. You found both roots of x² − 8x − 9 = 0, but x = −1 puts −1 inside a log. Every number inside a log must be above 0, so it must be rejected. Go to M5.2.4. Try another.

    2. Not this one. You may have taken 2 times 9. The product law gives x(x − 8) = 3² = 9, which is the quadratic x² − 8x − 9 = 0. Go to M5.2.4. Try another.

    3. Not this one. x = −1 is the root that does not work. It makes x negative, and the log of a negative number does not exist. Go to M5.2.4. Try another.

    4. Yes, that is right. The product law gives log₃ [x(x − 8)] = 2, so x(x − 8) = 9, and x² − 8x − 9 = (x − 9)(x + 1) = 0. We need x > 8, so x = −1 is rejected. Check: log₃ 9 + log₃ 1 = 2 + 0 = 2.

    See the answer

    x = 9 only

    The product law gives log₃ [x(x − 8)] = 2, so x(x − 8) = 9, and x² − 8x − 9 = (x − 9)(x + 1) = 0. We need x > 8, so x = −1 is rejected. Check: log₃ 9 + log₃ 1 = 2 + 0 = 2.

    • If you chose A: You found both roots of x² − 8x − 9 = 0, but x = −1 puts −1 inside a log. Every number inside a log must be above 0, so it must be rejected. Go to M5.2.4.
    • If you chose B: You may have taken 2 times 9. The product law gives x(x − 8) = 3² = 9, which is the quadratic x² − 8x − 9 = 0. Go to M5.2.4.
    • If you chose C: x = −1 is the root that does not work. It makes x negative, and the log of a negative number does not exist. Go to M5.2.4.

  4. Solve 0.5ˣ > 8.

    Tap your answer.

    1. Not this one. You did not flip the sign. Since 0.5ˣ gets smaller as x grows, the values above 8 are on the left, not the right. Go to M5.2.4. Try another.

    2. Yes, that is right. Since 0.5⁻³ = 8, the boundary is x = −3. The curve y = 0.5ˣ falls as x rises, so it is above 8 for x less than −3. With logs: x log 0.5 > log 8, and dividing by the negative log 0.5 flips the sign.

    3. Not this one. Check x = 4: 0.5⁴ = 0.0625, which is not above 8. A decay curve is above 8 only on the far left. Go to M5.2.4. Try another.

    4. Not this one. Check x = 0: 0.5⁰ = 1, which is not above 8. The boundary is at a negative x. Go to M5.2.4. Try another.

    See the answer

    x < −3

    Since 0.5⁻³ = 8, the boundary is x = −3. The curve y = 0.5ˣ falls as x rises, so it is above 8 for x less than −3. With logs: x log 0.5 > log 8, and dividing by the negative log 0.5 flips the sign.

    • If you chose A: You did not flip the sign. Since 0.5ˣ gets smaller as x grows, the values above 8 are on the left, not the right. Go to M5.2.4.
    • If you chose C: Check x = 4: 0.5⁴ = 0.0625, which is not above 8. A decay curve is above 8 only on the far left. Go to M5.2.4.
    • If you chose D: Check x = 0: 0.5⁰ = 1, which is not above 8. The boundary is at a negative x. Go to M5.2.4.

  5. From an earlier chapter

    Which of these is equal to 3 log 2 + log 5?

    Tap your answer.

    1. Not this one. You multiplied 3, 2 and 5. The 3 is a multiplier, so it becomes a power of 2: 3 log 2 = log 2³ = log 8. Go to M5.2.3. Try another.

    2. Not this one. You worked out 3 × 2 + 5. The sum of logs is the log of the product, not the log of the sum. Go to M5.2.3. Try another.

    3. Yes, that is right. The power law gives 3 log 2 = log (2³) = log 8. The product law gives log 8 + log 5 = log (8 × 5) = log 40.

    4. Not this one. You may have used the 3 as a power of both numbers: (2 × 5)³ = 1 000. The 3 belongs to the 2 only: log 8 + log 5 = log 40. Go to M5.2.3. Try another.

    See the answer

    log 40

    The power law gives 3 log 2 = log (2³) = log 8. The product law gives log 8 + log 5 = log (8 × 5) = log 40.

    • If you chose A: You multiplied 3, 2 and 5. The 3 is a multiplier, so it becomes a power of 2: 3 log 2 = log 2³ = log 8. Go to M5.2.3.
    • If you chose B: You worked out 3 × 2 + 5. The sum of logs is the log of the product, not the log of the sum. Go to M5.2.3.
    • If you chose D: You may have used the 3 as a power of both numbers: (2 × 5)³ = 1 000. The 3 belongs to the 2 only: log 8 + log 5 = log 40. Go to M5.2.3.

  6. Solve 3ˣ = 50. Give x to 3 significant figures.

    See the answer

    3.56

    Take logs: x log 3 = log 50. So x = log 50 ÷ log 3 = 1.69897 ÷ 0.47712 = 3.5609, which is 3.56 to 3 significant figures.

  7. A culture has 500 cells and the number triples every hour. After how many whole hours does it first have more than 1 000 000 cells?

    See the answer

    7

    After n hours there are 500 × 3ⁿ cells. We need 500 × 3ⁿ > 1 000 000, so 3ⁿ > 2 000. Take logs: n > log 2 000 ÷ log 3 = 6.92. The smallest whole number is 7. Check: 500 × 3⁷ = 1 093 500, and 500 × 3⁶ = 364 500.

  8. Solve log₃ (2x + 1) − log₃ x = 2. Give x as a fraction.

    See the answer

    1/7

    The quotient law gives log₃ [(2x + 1) ÷ x] = 2. In index form, (2x + 1) ÷ x = 3² = 9. So 2x + 1 = 9x, which gives 1 = 7x and x = 1/7. Check: 2x + 1 = 9/7, and (9/7) ÷ (1/7) = 9, with log₃ 9 = 2.

  9. Solve 4ˣ⁺¹ − 17(2ˣ) + 4 = 0.

    Tap your answer.

    1. Not this one. These are the values of u = 2ˣ. You must go back to x: 2ˣ = 4 gives x = 2, and 2ˣ = 1/4 gives x = −2. Go to M5.2.4. Try another.

    2. Not this one. The root u = 1/4 is above 0, so it gives a solution too: 2ˣ = 1/4 gives x = −2. Go to M5.2.4. Try another.

    3. Not this one. You went back with base 4. But u = 2ˣ, so 2ˣ = 1/4 gives x = −2 and 2ˣ = 4 gives x = 2. Go to M5.2.4. Try another.

    4. Yes, that is right. Split the power: 4ˣ⁺¹ = 4 × 4ˣ = 4(2ˣ)². With u = 2ˣ the equation is 4u² − 17u + 4 = 0, which is (4u − 1)(u − 4) = 0. So u = 1/4 or u = 4, giving x = −2 or x = 2.

    See the answer

    x = −2 or x = 2

    Split the power: 4ˣ⁺¹ = 4 × 4ˣ = 4(2ˣ)². With u = 2ˣ the equation is 4u² − 17u + 4 = 0, which is (4u − 1)(u − 4) = 0. So u = 1/4 or u = 4, giving x = −2 or x = 2.

    • If you chose A: These are the values of u = 2ˣ. You must go back to x: 2ˣ = 4 gives x = 2, and 2ˣ = 1/4 gives x = −2. Go to M5.2.4.
    • If you chose B: The root u = 1/4 is above 0, so it gives a solution too: 2ˣ = 1/4 gives x = −2. Go to M5.2.4.
    • If you chose C: You went back with base 4. But u = 2ˣ, so 2ˣ = 1/4 gives x = −2 and 2ˣ = 4 gives x = 2. Go to M5.2.4.

  10. Solve log₃ x = log₉ (x + 20).

    Tap your answer.

    1. Yes, that is right. Since 9 = 3², log₉ (x + 20) = 1/2 log₃ (x + 20). So 2 log₃ x = log₃ (x + 20), which gives x² = x + 20, or (x − 5)(x + 4) = 0. We need x above 0, so x = 5. Check: log₃ 5 = 1.465 and log₉ 25 = 1.465.

    2. Not this one. You may have halved the number inside instead of the log: x = (x + 20) ÷ 2 gives 20. The change of base halves the log: log₉ (x + 20) = 1/2 log₃ (x + 20), so x² = x + 20. Go to M5.2.4. Try another.

    3. Not this one. The root x = −4 is the one that fails. It puts −4 inside log₃ x. Go to M5.2.4. Try another.

    4. Not this one. The root x = −4 puts −4 inside log₃ x, and the log of a negative number does not exist. It must be rejected. Go to M5.2.4. Try another.

    See the answer

    x = 5

    Since 9 = 3², log₉ (x + 20) = 1/2 log₃ (x + 20). So 2 log₃ x = log₃ (x + 20), which gives x² = x + 20, or (x − 5)(x + 4) = 0. We need x above 0, so x = 5. Check: log₃ 5 = 1.465 and log₉ 25 = 1.465.

    • If you chose B: You may have halved the number inside instead of the log: x = (x + 20) ÷ 2 gives 20. The change of base halves the log: log₉ (x + 20) = 1/2 log₃ (x + 20), so x² = x + 20. Go to M5.2.4.
    • If you chose C: The root x = −4 is the one that fails. It puts −4 inside log₃ x. Go to M5.2.4.
    • If you chose D: The root x = −4 puts −4 inside log₃ x, and the log of a negative number does not exist. It must be rejected. Go to M5.2.4.

  11. Solve (log₃ x)² − log₃ x − 2 = 0.

    Tap your answer.

    1. Not this one. These are the values of y = log₃ x. You must go back to x: 3² = 9 and 3⁻¹ = 1/3. Go to M5.2.4. Try another.

    2. Yes, that is right. Let y = log₃ x. Then y² − y − 2 = 0, which is (y − 2)(y + 1) = 0. So y = 2 or y = −1. Then x = 3² = 9 or x = 3⁻¹ = 1/3. Both are above 0.

    3. Not this one. You may have taken y = −1 to mean x = −3. A power of 3 is never negative: y = −1 gives x = 3⁻¹ = 1/3. Go to M5.2.4. Try another.

    4. Not this one. The root y = −1 is allowed, because log₃ x can be negative. It gives x = 3⁻¹ = 1/3. Go to M5.2.4. Try another.

    See the answer

    x = 9 or x = 1/3

    Let y = log₃ x. Then y² − y − 2 = 0, which is (y − 2)(y + 1) = 0. So y = 2 or y = −1. Then x = 3² = 9 or x = 3⁻¹ = 1/3. Both are above 0.

    • If you chose A: These are the values of y = log₃ x. You must go back to x: 3² = 9 and 3⁻¹ = 1/3. Go to M5.2.4.
    • If you chose C: You may have taken y = −1 to mean x = −3. A power of 3 is never negative: y = −1 gives x = 3⁻¹ = 1/3. Go to M5.2.4.
    • If you chose D: The root y = −1 is allowed, because log₃ x can be negative. It gives x = 3⁻¹ = 1/3. Go to M5.2.4.

  12. Solve log₂ x + log₂ y = 5 and log₂ x − log₂ y = 1 together.

    Tap your answer.

    1. Not this one. You solved x + y = 5 and x − y = 1. The equations are about logs: the first gives xy = 2⁵ and the second gives x ÷ y = 2¹. Go to M5.2.4. Try another.

    2. Not this one. You swapped x and y. The second equation says x ÷ y = 2, so x is the bigger one. Go to M5.2.4. Try another.

    3. Not this one. The value y = −4 puts −4 inside log₂ y. A number inside a log must be above 0. Go to M5.2.4. Try another.

    4. Yes, that is right. The product law gives log₂ (xy) = 5, so xy = 32. The quotient law gives log₂ (x ÷ y) = 1, so x ÷ y = 2 and x = 2y. Then 2y² = 32, so y = 4 (not −4). So x = 8.

    See the answer

    x = 8, y = 4

    The product law gives log₂ (xy) = 5, so xy = 32. The quotient law gives log₂ (x ÷ y) = 1, so x ÷ y = 2 and x = 2y. Then 2y² = 32, so y = 4 (not −4). So x = 8.

    • If you chose A: You solved x + y = 5 and x − y = 1. The equations are about logs: the first gives xy = 2⁵ and the second gives x ÷ y = 2¹. Go to M5.2.4.
    • If you chose B: You swapped x and y. The second equation says x ÷ y = 2, so x is the bigger one. Go to M5.2.4.
    • If you chose C: The value y = −4 puts −4 inside log₂ y. A number inside a log must be above 0. Go to M5.2.4.

Keep it

Copy this into your exercise book.

aˣ = b: take logs, x = log b ÷ log a. Disguised quadratic: let u = aˣ, keep u > 0. Log equations: one log, index form, reject answers that make a number inside a log 0 or less.

Draw: The curve y = 3ˣ, the dashed line y = 20, and the point where they meet with its guide lines.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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